Secant varieties
Marin Petkovic
November 23, 2015
Abstract
The goal of this talk is to introduce secant varieies and show connections of secant varieties of Veronese variety to the Waring problem.
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Secant varieties
Let X ⊂ Pn be a variety over k, where k is algebraically closed field of characteristic 0 (or just assume k = C).
Definition 1.1. For s ≥ 1 the s-th higher secant variety of X is
σs (X) =
[
hP1 , . . . , Ps i.
P1 ,...,Ps ∈X
Remark 1.2. Although in some cases closure is not necessary, for twisted cubic
in P3 the union of secants is not closed.
Lemma 1.3. If σi (X) = σi+1 (X), then σi (X) is a linear subspace of PN .
S
Proof. Let z ∈ P1 ,...,Pi ∈X hP1 , . . . , Pi i be nonsingular point on σi (X). Then
hz, xi ⊂ σi+1 for each x ∈ X. Therefore hz, xi ⊂ Tz (σi+1 ) = Tz (σi ). Then
σi ⊂ Tz (σi ). Since they have the same dimension, σi = Tz (σi ), i.e. σi is
linear.
Proposition 1.4. If X is nondegenerate (not contained in a hyperplane), then
there exists r ≥ 1 such that
X = σ1 (X) ( σ2 (X) ( . . . ( σr (X) = PN .
Now we can ask two natural questions about higher secant varieties. First,
what is the minimal r such that σr (X) = PN ? More generally, what is the
dimension dim(σi (X))?
Proposition 1.5.
dim(σi (X)) ≤ min{in + i − 1, N }.
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We will call the number min{in + i − 1, N } the expected dimension of σi (X)
and denote it with expdim(σi (X)). If dim(σi (X)) 6= expdim(σi (X)), we say
that X is i-defective.
Remark 1.6. If X is i-defective and σi (X) 6= PN , then X is j-defective, for
all j ≤ i, because of dim(σi+1 (X)) ≤ dim(σi (X)) + n + 1.
Example 1.7 (Veronese surface). Let X be the Veronese surface, that is the
image of the map
v2 : P 2 → P 5
(x : y : z) 7→ (x2 : xy : xz : y 2 : yz : z 2 ).
The expected dimension of σ2 (X) is 5, but we will show that σ2 (X) 6= P5 , so X
is 2-defective.
To prove this, we identify elements of P5 with 3 × 3 symmetric matrices.
Notice that the elements of X are symmetric matrices of rank 1. Thus σ2 (X)
consists of all linear combinations of 2 rank 1 symmetric matrices, thus of rank
≤ 2. Hence σ2 (X) 6= P5 , i.e. X is 2-defective.
Lemma 1.8 (Terracini’s lemma). Let P1 , . . . , Ps ∈ X be general points and
P ∈ hP1 , . . . , Ps i ⊂ σs (X) be a general point. Then the tangent space to σs (X)
in P is
TP (σs (X)) = hTP1 (X), . . . , TPs (X)i.
Example 1.9 (Twisted cubic curve). Let X ⊂ P3 be a twisted cubic curve, that
is the image of the map v3 : P1 → P3 defined with
v3 (x : y) = (x3 : x2 y : xy 2 : y 3 ).
Then dim(σ2 (X)) = dim(TP (σ2 (X))) for generic point P .
lemma, we have
TP (σ2 (X)) = hTP1 (X), TP2 (X)i
By Terracini’s
for some points P1 , P2 ∈ X. Since in general case TP1 (X), TP2 (X) do not
intersect, they span 3-dimensional space, that is P3 .
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Veronese varieties and the Waring problem
Definition 2.1. The d-th Veronese map vd : Pn → PN , where N =
is defined with
n+d
d
− 1,
vd (x0 : . . . : xn ) = (xi00 xi11 · · · xinn |i1 + . . . + in = d)
The image of vd is called the Veronese variety.
Remark 2.2. Veronese map can also be defined as a map taking linear form
L to Ld . Taking appropriate basis and identifying P1 with the space of linear
forms this map corresponds to vd in the above definition. We will denote the
point corresponding to F with [F ]. Notice that [F ] ∈ PN is in X if and only if
F = Ld for some linear form L.
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Proposition 2.3. The tangent space to the Veronese variety at the point [Ld ]
is
h [Ld−1 M ] ; M linear form i.
Example 2.4. If n = 1, we call the Veronese varieties rational normal curves.
The curve v2 (P1 ) ⊂ P2 is just the irreducible conic, so σ2 (v2 (P1 ) = P2 . The
curve v3 (P1 ) ⊂ P3 is the twisted cubic curve.
The higher secant varieties of the Veronese variety are connected with the
Waring problem for forms, which we will introduce now.
Definition 2.5. Let F be a degree s homogeneous form. The Waring rank of
F is the minimum s such that
F = Ld1 + . . . + Lds
for some linear forms Li . We denote the Waring rank of F with rk(F ).
The big Waring problem: Find the minimal integer g(n, d) such that
rk(F ) ≤ g(n, d),
for a generic homogeneous form F of degree d in n + 1 variables.
The little Waring problem: Find the minimal integer G(n, d) such that
rk(F ) ≤ G(n, d),
for all homogeneous forms F of degree d in n + 1 variables.
Remark 2.6. It is clear that solving the big Waring problem is equivalent to
finding minimal s such that σs (X) = PN for Veronese variety X = vd (Pn ) ⊂
PN . This is not enough to solve the little Waring problem, because of the closure
in the definition of secant varieties.
Example 2.7. Let X = v2 (P1 ) ⊂ P2 be an irreducible conic. Then σ2 (X) = P2 ,
so g(1, 2) = 2. Moreover, since in this case the closure is not necessary in the
definition of σ2 (X), we also get G(1, 2) = 2.
Example 2.8. Let X = v3 (P1 ) ⊂ P3 be the twisted cubic curve. Then σ2 (X) =
P3 so g(1, 3) = 2. But in this case, there are forms with rank 3. We can explain
this geometrically. Let [F ] be a point in P3 corresponding to F . Let π be the
projection of X from the point [F ]. Then the image is the plane cubic curve
which has to be singular. If the singularity is a node, then [F ] lies in on a
secant of X, so F = L31 + L32 . But if singularity is a cusp, then [F ] lies on a
tangent line of X, and not on a secant, and rk(F ) = 3.
Let P1 , . . . , Ps ∈ X = Pn be distinct points. Then each hyperplane H
containing all tangent spaces Tvd (Pi ) (X) corresponds to a degree d hypersurface
in Pn that is singular in points P1 , . . . , Ps . Such hypersurfaces are defined by a
homogeneous degree d form contained in p21 ∩ . . . ∩ p2s , where pi is the ideal of
the point Pi . Thus if (p21 ∩ . . . ∩ p2s )d is not empty, σs (X) 6= PN . Moreover, since
two different hypersurfaces give two different hyperplanes, we have the following
result:
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Lemma 2.9. Let P1 , . . . , Ps ∈ Pn be generic points. Then
dims (σ(X)) = N − dim(p21 ∩ . . . ∩ p2s )d .
From the expected value for dim(σs (X)) we can easily calculate the expected
value for g(n, d) to be
&
'
n+d
d
n+1
.
Theorem 2.10 (Alexander and Hirtschowitz). Let F be a generic degree d form
in n + 1 variables. Then
&
'
n+d
d
rk(F ) =
n+1
except in the following cases:
• d = 2, where rk(F ) = n + 1
• d = 4, n = 2, where rk(F ) = 6
• d = 4, n = 3, where rk(F ) = 10
• d = 3, n = 4, where rk(F ) = 8
• d = 4, n = 4, where rk(F ) = 15.
It is easy to explain the case d = 2 using the symmetric matrices. Indeed, a
general n + 1 × n + 1 matrix is regular, that is of rank n + 1. Hence it is a sum
of n + 1 matrices of rank 1. Therefore the rank of associated form is n + 1.
Example 2.11 (case (d = 4, n = 2)). Let X = v4 (P2 ) ⊂ P14 . We prove that
σ5 (X) 6= P14 . By the lemma and the discussion before, it is enough to show that
there exists a degree 4 hypersurface through 5 generic points that is singular in
those 5 points.
Example 2.12 (case d = 4, n = 3, 4). To show that g(4, 3) > 9 and g(4, 4) > 14
it is sufficient to find a quadric hypersurface in P3 through 9 given points, and
a quadric hypersurface in P4 through 14 points.
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Apolarity
Let S = k[x1 , . . . , xn ] and T = k[y1 , . . . , yn ]. We define the action of T on S
with
∂
yi .xj =
xj
∂xi
and extend it to T .
For a graded ring R, we denote with Rd the d-th graded subring.
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Definition 3.1. For F ∈ S homogenous we define the anihilator with
F ⊥ = {∂ ∈ T | ∂F = 0}.
Definition 3.2. Let V, W be finite dimensional k-vector spaces and B : V ×
W → k a bilinear map. We say that B is a perfect pairing if the induced maps
V → W ∗ and W → V ∗ are isomorphisms.
Proposition 3.3. The bilinear map Sd × Td → k defined with (F, ∂) 7→ ∂F is
a perfect pairing.
Definition 3.4. For an ideal I ⊂ T we define the Hilbert function of T /I with
HF (T /I, t) = dim(T /I)t ,
for t ≥ 0.
In the case I = F ⊥ we will use the notation HF (t) instead HF (T /F ⊥ , t).
Let m be the maximal ideal (y1 , . . . , yn ) ⊂ T . We define the sockle of the
ring (T /F ⊥ ) with
Soc(T /F ⊥ ) = {x ∈ T /F ⊥ k xm = 0}.
It is an ideal in T /F ⊥ and it is easy to check
(
0
⊥
dim Soc(T /F )i =
1
if i 6= d
if i = d
Proposition 3.5. Hilbert function of F ⊥ for F of degree d vanishes for t > d,
HF (0) = HF (d) = 1, and is symmetric with respect to d+1
2 .
Proof. The first part is obvious since ∂F = 0 for each ∂ of degree > d. From the
perfect pairing property it follows that not all ∂ ∈ Td vanish in F , so HF (d) = 1.
It is obvious that HF (0) = 1.
Let R = T /F ⊥ ). To prove that HF is symmetric, it is enough to prove that
the multiplication Ri × Rd−i → Rd = k is perfect pairing. Let x ∈ Ri such that
xy = 0 for each y ∈ Rd−i . We will prove that x is in Soc(R). If y ∈ Rl for
0 < l < d − i, using downward induction we have xym = 0, that is xy ∈ Soc(R).
Since deg(xy) < d this is a contradiction. Thus y ∈ Soc(R), so y = 0.
Example 3.6. Let F ∈ k[x0 , x1 ] be homogenous cubic. We will compute all
the possible Waring ranks of F . The Hilbert function of F is determined by the
value for t = 1, so there is only two possibilities: HF (1) = 1 or HF (1) = 2.
Notice that since (F ⊥ )1 = 0 rank of F cannot be 1.
Assume the first case. Then (F ⊥ )1 = h∂1 i. The space of all linear forms L
such that ∂1 L = 0 is not S1 so it is one dimensional, that is equal to hL1 i for
some L1 . Thus we can take L0 such that ∂1 L0 = 1, and S1 = hL0 , L1 i. Then
there exists a polynomial G such that
G(L0 , L1 ) = aL30 + bL20 L1 + cL0 L21 + dL31 = F (x0 , x1 ).
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Since ∂1 (L0 ) = 1 and ∂1 (L1 ) = 0 we get
0 = ∂1 G(L0 , L1 ) = 3aL20 + 2bL0 L1 + cL21 ,
so F (x0 , x1 ) = dL31 , and rk(F ) = 1.
In the second case, we have (F ⊥ )2 = hQi. Then Q = ∂∂ 0 , where ∂, ∂ 0 ∈ T1 .
Suppose that ∂, ∂ 0 are linearly independent. We can find a basis L, L0 for S1
such that ∂L = ∂ 0 L0 = 1 and ∂L0 = ∂ 0 L = 0. Then we take polynomial G as
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before and Show that G(L, L0 ) − aL3 − dL0 is a zero polynomial (∂G = 0 for
all ∂ ∈ T3 ). Thus rk(F ) = 2.
Suppose now that Q = ∂ 2 and that F = N 3 + M 3 , that is rk(F ) = 2. Then
we can take ∂N , ∂M such that ∂M M = ∂N N = 1 and ∂M N = ∂N M = 0. Then
∂M ∂N ∈ F ⊥ and that is a contradiction since (F ⊥ )2 = (Q).
Notice that in the previous example for the forms of rank 1, F ⊥ ⊃ (∂), and
(∂) is an ideal of a point in P1 . Similar, in the case of rank 2, F ⊥ ⊃ (∂∂ 0 ) which
is an ideal of two points in P1 . Since (F ⊥ )1 = 0 there is no ideals of one point
in F ⊥ . Also, in the case of rank 3, there are no ideals of one or two points in
F ⊥ , but there are ideals of three points.
Lemma 3.7 (Apolarity Lemma). F = Ld1 + . . . + Lds ⇐⇒ F ⊥ contains an ideal
of s distinct points.
Remark 3.8. If I ⊂ F ⊥ , then obviously HF (T /I, t) ≥ HF (t).
Example 3.9. For F = x0 x21 we see that F ⊥ = (∂02 , ∂13 ), so it does not contain
an ideal of one or two points. Therefore rk(F ) = 3 ((∂1 ∂02 + ∂13 ) is an ideal of
3 points). More generally, we can conclude that rk(x0 xd1 ) = d + 1.
Example 3.10 (defective cases for Alexander Hirowitz theorem). Let F be a
generic form of degree 4, n = 2. The Hilbert function is given with HF (1) = 3,
HF (2) = 6, that is, no first or second order differentials vanish in F . If I is an
ideal of five points, we have HF (T /I, 2) = 5 (i.e. I contains a quadric), so F ⊥
contains no ideals of five points.
Similarly, in the case n = 3 we have HF (2) = 2+3
= 10 > 9, and
2
⊥
HF (T /I, 2) = 9 for and ideal I of nine points.
Thus
I
⊂
6
F
.
In the case n = 4 we have HF (2) = 2+4
= 15 > 14 so F ⊥ does not contain
2
an ideal of 14 points.
References
[1] Enrico Carlini, Nathan Grieve, and Luke Oeding, Four Lectures on Secant
Varieties, Volume 76 of the series Springer Proceedings in Mathematics &
Statistics pp 101-146, 2014.
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